Final answer:
The chemical reaction of aqueous bromine with propane in sunlight or aqueous bromine with propene without sunlight can be explained using the concept of halogenation. In the presence of light or heat, alkanes react with halogens to yield halogenated alkanes. The reaction equations for these reactions involve the substitution of hydrogen atoms with bromine atoms to form brominated products.
Step-by-step explanation:
The chemical reaction of aqueous bromine with propane in sunlight or aqueous bromine with propene without sunlight can be explained using the concept of halogenation. In halogenation reactions, alkanes react with halogens in the presence of light or heat to yield halogenated alkanes.
The reaction between aqueous bromine and propane in sunlight can be represented by the following balanced equation:
C₃H₈ + Br₂(C₃H₇Br) + HBr
Here, the bromine molecule (Br₂) reacts with propane (C₃H₈) to form 2-bromopropane (C₃H₇Br) and hydrogen bromide (HBr).
On the other hand, in the absence of sunlight, the reaction between aqueous bromine and propene can be represented by the following balanced equation:
C₃H₆ + Br₂(C₃H₅Br) + HBr
In this case, the bromine molecule reacts with propene (C₃H₆) to form 1-bromopropane (C₃H₅Br) and hydrogen bromide (HBr).